Discussion
Find the average of first '\(n\)' natural numbers
Answer: C
The average is \(\frac{n+1}{2}\).
The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\); dividing by the number of terms, \(n\), gives the average:
\[ \frac{1}{n} \times \frac{n(n+1)}{2} = \frac{n+1}{2} \]
For example, the average of 1 to 5 is 3 = (5 + 1)/2. \(\frac{n(n+1)}{2}\) is the sum, not the average; \(\frac{n(n+1)(2n+1)}{2}\) is a garbled form of the sum of squares, and \(n^{2}\) is the sum of the first \(n\) odd numbers.
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